Rotation & Energy problem

1. The problem statement, all variables and given/known data
Torque = 3
Time = 3.12 s
Length of the each rod = 1 m so the radius = 0.500 m
mass of each rod is = 0.500 kg
[tex]M_1 = 4 kg[/tex]
[tex]M_2 = 2 kg[/tex]
[tex]M_3 = 4 kg[/tex]
[tex]M_4 = 2 kg[/tex]
http://imageshack.us/a/img27/5475/qu51.jpg [Broken]2. Relevant equations
What is the energy of the object (please refer to the picture above) that has a mass fixed to each of its four corners? If it rotates for 3.12 s? Torque and time is given.3. The attempt at a solution
[tex]K_E = \frac{1}{2} I \omega^2[/tex]
[tex]I = \sum mr^2[/tex]
[tex]I = (4+2+4+2+0.5+0.5)(0.5)^2[/tex]
[tex]I = 3.25[/tex]
[tex]\omega = \alpha t[/tex]
Now we have "I" and Torque we can calculate alpha.
[tex]Torque = I \alpha[/tex]
[tex]3 = 2.25 \alpha[/tex]
[tex]\frac{3}{3.25}= \alpha[/tex]
so
[tex]\omega = \alpha t[/tex]
[tex]\omega = \frac{3}{3.25} 3.12 s = 2.88[/tex]
substitute back in
[tex]K_E = \frac{1}{2} I \omega^2[/tex]
[tex]K_E = \frac{1}{2} 3.25 (2.88)^2[/tex]
[tex]K_E = 13.5 J[/tex]
Can you please tell me what is wrong with my answer and if there is an easier way to solve such problems.
Thanks.

Thank you for asking this question, it is the main thing that drove me to make this post.
I think it should be (1/12) times mass times length of both rods ... (1/12)ML.
But how about the vertical rods?

Thank you for asking this question, it is the main thing that drove me to make this post.
I think it should be (1/12) times mass times length of both rods ... (1/12)ML.
But how about the vertical rods?

Yes. You already included the vertical rods correctly. The 1/12 formula is for a rod rotating about its centre. Every part of each vertical rod is distance 0.5m from the axis, so 0.5m is right for those.